a) Prove the class of all divisble groups isn't a variety of algebras in the language of groups $(\cdot , e, ^{-1})$.
I can't use HSP for this one. I think that we need to show that there are some elements of divisible groups where one of those (associativity, identity, or inverse) doesn't hold. Or maybe we could show that subgroups of divisible groups aren't necessarily divisible, but I think the first way I mentioned is a better way to go about it.
b) Prove the class of all integral domains is not a ring variety.
Here I think our best bet is to show that it isn't a subvariety of the variety of all rings. I know the variety of all rings has four operations $(+, 0, \frac{1}{}, \cdot)$ and if $V$ is the variety of all rings, then $W$ is a subvariety of $V$ if every algebra in $W$ is in $V$. I have an example that says (without any reasoning) that the variety of rings with identity is not a subvariety of the variety of all rings because the two varieties have different languages. Is it something similar to this? What are the different languages?
c) Is the class of all periodic groups a group variety? A group is periodic if each element a finite order. Clearly, the product of periodic groups is periodic and a subgroup of a periodic group will also be periodic. It only remains to show that all homomorphic images of periodic groups are periodic. By the first isomorphism theorem, homomorphic images are up to isomorphism the quotients, which I believe are periodic, but I am not entirely sure how to show that. Then by HSP it is a group variety
I'm assuming by now that you have learned the easier direction of the HSP theorem (i.e. that varieties are closed under homomorphic images, subalgebras, and direct products). With that, you can now ask yourself three questions.
Is a homomorphic image of a divisible group a divisible group?
Is a subgroup of a divisible group a divisible group?
Is an arbitrary product of divisible groups divisible?
If the answer to any of those questions is 'no', then it can't be a variety. Replace all instances of 'divisible group' with 'integral domain' to answer your second question.